A window through the walls of our classroom. This is an interactive learning ecology for students and parents in my Calculus 45S class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Monday, April 28, 2008

Why hello there, it's Kristin_R here, one of the few non-AP students left in the calculus class.

Today in class we learned about the first and second derivative tests.

The first derivative test:

c is a critical point (meaning the inputs where the derivative is zero or undefined, which is a horizontal tangent line) somewhere over the interval (a,b).

If f'(x) > 0 over the interval (a,c) and f'(x) < 0 for all values of x in the interval (c, b), then that function has a maximum over that interval.

If f'(x) ,0 over the interval (a,c) and f'(x) . 0 over the interval (c,b), then that function has a minimum over that interval.

Here's an example...

f(x) = x^2-8x=4

therefore, f'(x) = 2x-8

If we let the derivative equal zero, the root is at x=4.

When we plug four back into the original function, we get a y-value of -12. .

This means that the root is at (4, -12).

- ----- -12 ------ - --------4-------- +

By the first derivative test, x=4 is a minimum because to the left of 4, f is decreasing , and to the right of four, f is increasing.

The second derivative test:

When the parent function is concave up (valley), f'' is postive.

When the parent function is concave down (hill), f'' is negative.

When f' has a root, the parent function has a point of inflection, which is where the function changes concavity.

So, the first derivative test shows where the function is increasing or decreasing, as well as maximum or minimum extrema.

The second derivative test shows us the concavity of the graph.

"Just finding a root of your second derivative is not enough! You have to check your intervals!"

Just a quick little note - in grade eleven, you should have been taught about how to graph certain expontential functions. Remember that the degree of the given function is the same amount of possible roots in the graph.

So, thats it, thats the whole ball of wax...y'all dig?

Homework is up to and including page 64 in the white and orange book, as well as up to and including page 42 in the blue book.

Sorry it's so late. Kept forgetting and procrastinating. Yes, it was supposed to be me... but, I wasn't even participating during the class, because of APcalc in the corner of the room. But, I can at least explain the slides.

Slide 1, 2

Picture and the question.

Slide 3,4,5

Velocity is the change of distance over the change in time.

a. The average velocity, is the slope of the line, from the interval 0 to 4 of the given position function.

b. Velocity is zero when, the derivative of x(t) is equal to zero.

c. When the point is moving to the right (positive) meaning, the velocity is positive. (Derivative of x is positive)

d. Same thing as c, but moving left (negative)

e. When the derivative is equal to 3

f. Taking the derivative of the velocity function. Or the second derivative of the position function, called the acceleration function is equal to 3.

The graph, and line test means, you are trying to find from the derivative, where there may be local minimums or maximums on the parent function.

So, when the derivative has values that crosses the x axis(zero) there is a local maximum or minimum. To determine whether it is a maximum or a minimum, on the left side of the zero, if it is a positive value then crosses the x-axis and then is negative, it is a local maximum. Vice versa, it is a local minimum.

Slide 8

The extreme value theorem, says, on a differentiated function, within a closed interval, there exists a maximum and minimum value on the parent function.

Sunday, April 27, 2008

In this episode Timothy came back to school on Friday afternoon to talk about his week attending the miniUniversity program at the University of Winnipeg. He talks about the differences he finds between teaching and learning at high school and university and describes learning in the university classroom using a thought provoking metaphor, listen for it. Also, we have a cameo appearance by two very special people at the very end.

Monday, April 21, 2008

On Friday, our class started talking about the application of derivatives. To start off, Mr. K showed us this slide:

Mr. K then asked us when was the car moving fastest. The class decided that the car was moving fastest during t2, because the slope of the tangent line was the steepest at that point. During the interval [t1,t2], it is clear that the car is speeding up during that interval because the slope of the tangent lines are getting steeper going from t1 to t2. The car is definitely slowing down on the interval [t2,t3] as we can see that the slope of the tangent lines are decreasing.

Now this was a tough graph to decipher. The AP Calculus students had no problem understanding this one, but to those of us who aren't taking AP, this was kind of difficult. I myself found it quite confusing at first, but after clarification, I understand it. Now, you have to find the derivative of f in this graph to find out where is f ' is positive. First we have to find out where the tangent lines in f are zero. Looking at the function f, we can clearly see that the tangent lines are zero in -2 and +2. This means that f ' will have roots at -2 and +2. Then, we look when the tangent lines of f are approaching zero. There are 2 instances in this graph. The first instance is from (-∞, -2]. The slope from that domain is increasing, since it's coming from a negative value to zero. But that still doesn't solve our problem since f ' is still negative at that point. Then, from (-2, 2) in f, we can see that the slope is increasing until (0,0) and it starts to decrease until it reaches (2,3), where it becomes zero. At this point, f ' is positive as the slopes of the tangent lines from (-2, -1) to (0,0) are increasing from zero and reaches its maximum at (0,0), and then it starts to decrease from (0,0) until it reaches (2,3) where the slope is zero. This then solves the first question which asks where is f ' positive, which is from (-2, o) to (2,0).

Now to find out where f '' is positive. From the graph, we can see that f is some kind of a cubic function, so we know that its derivative will be a quadratic function. Now, the derivative of a quadratic function is a straight line, so now we know that f '' is a straight line. Looking at the shape of f ' (say it is -2x2), it is definitely a negative quadratic function because it has a maximum, not a minimum. We then have to find out the derivative of f '' which is represented as -2x2 (However, it IS NOT -2x2, it's just a representation). Using the power rule, we can determine that the derivative of f ', which is f '', is -4x. From this we can see that the function is positive at quadrant 2.

The critical numbers (which is another term for the roots) of f are -2 and 2.

This is a little difficult...For a), f is increasing wherever f ' is positive.For b), f(0) is negative.

Well, that was all we talked about last class. Mr. K tried to squeeze in one more slide but we ran out of time.

Saturday, April 19, 2008

I was talking to Jessie, one of my Applied Math students, earlier this week while helping her review over the lunch hour. I found her comments so compelling I asked her (and later her parents) if I could record and publish her comments so other students could hear what she had to say. I've long thought students need to hear from other students how they best learn to help them all learn.

This is the first in what I hope will be a series of podcasts called Student Voices. I'm hoping to have one of these short conversations with a student published each week. If you'd like to volunteer to be featured in one of these just let me know.

In this episode Jessie shares how she uses her class blog to learn and describes her personal "tipping point" from being confused to understanding Statistics very well. She also discusses the value of learning conversations and how sometimes being a "teacher" and sometimes a student helps her learn.

Wednesday, April 16, 2008

Something that I found that I thought was cool was when Mr. K showed us of something "brilliant" where we added terms - like adding zero. This unit I thought was funner than the first unit - Limits. When you get to cancel or reduce terms and create a whole new equation different from the beginning it creates a feeling of completion if it is the derivative of that equation. I feel that I need some more work towards this unit to fully understand it - overall I like this unit out of the two: Limits and Derivatives.

Tuesday, April 15, 2008

Anyway, I thought this unit was pretty fun and stuff. There are some things I still have to work on, which is knowing when to apply the product rule and stuff. Other than that I think I'm pretty much know what's going on. A good sleep and breakfast should do the trick, and a good review too.

Well, as I sit here and watch the hockey game, I will take some time (during an intermission of course) to reflect upon our last unit:

DERIVATIVES

Now I have already learned the content of this unit, and applied it many a time in the AP Calculus class. This means that everything on the test should be fairly easy for myself, but then again, it is one of Mr. K.'s tests.As I saw with the Pre-Test, small mistakes are easy to make. Forgetting to take the derivate of a constant, or the other side of an equation can result in a major headache.I think a good sleep, a hearty breakfast, and a little extra concentration should do the trick.

Good luck to all on the test.

Remember the derivative rules as well as the limit definition of a derivative!

Hey, MrSiwWy here for Monday's scribe! Now, all that we really did during class was write the pre-test, though Graeme, Mark and I all decided to rush through it and discuss it in our own group so that we could leave early to practice for the Literacy presentation in the afternoon. I know that it was my fault that we lost a few silly marks because of the fact that we rushed through everything on Monday, but since I really haphazardly rushed and finished way too early some of my solutions were flawed (and we ended up handing in my paper). Well, that's in the past, so on to the scribe. Now, usually a scribe for a pre-test class wouldn't really be that detailed, but I feel obligated to try and explain the solution process for each of the questions on the pre-test as well as I can. I'll accompany each with the corresponding slide, to show Mr. K's work (if that question was brought up in class) and to show the question itself.

1. What is dc/dx, where c is a constant?

Well, this is simply testing to see if you remember the constant rule for differentiation. Thus the answer is simply (b) which is 0. It might help if I reviewed with you the basic idea of each of the rules we have learned for differentiation, since the test is just about 12 hours away from now.Constant Rule: The constant rule basically dictates that any function that is solely just a constant, such as f(x) = 4, or g(x) = 13, has a derivative that is 0, or f'(x) = 0 and g'(x) = 0.Coefficient Rule: When differentiating a first-order function with a coefficient, such as f(x) = 2x, or g(x) = 215x, the derivative will simply be that coefficient. In the examples given, f(x) = 2x yields f'(x) = 2, while g(x) = 215x yields g'(x) = 215.Sum/Difference Rule: When a function is composed of more than one term, then you can break that function up into it's constituent parts and differentiate each part separately. Examples include f(x) = 388x + 6, or g(x) = 56x2 - 21x, where f'(x) = 388 + 0 = 388, or g'(x) = 102x - 21.Power Rule: For any function where x is raised to an exponent n, or f(x) = xn, the derivative of that function will be x raised to the exponent (n-1) all multiplied by n, or f'(x) = nxn-1.Product Rule: For any function that is composed of two functions being multiplied together, the derivative is not the derivative of their product, nor is it the product of their respective derivatives. To find the derivative of a product, the derivative will always be given by h'(x) = f(x)g'(x) + g(x)f'(x), assuming h(x) = f(x)g(x). It's basically the derivative of the first function multiplied by the other function, then add on the derivative of the second function multiplied by the first function.Quotient Rule:To find the derivative of a quotient, all you have to do is recall the quotient rule song! Assuming that h(x) = f(x) / g(x) then h'(x) = [g'(x)f(x) - f'(x)g(x)]/g2(x). Here's the song in case you forgot: "High de low minus low dehigh, all over low low"Chain Rule: This must be used whenever a function is actually buried deep inside another function. This can be given by h(x) = f(g(x)), such as h(x) = (2x2 + 3x)3. First you must find what g(x) is, or what the inner function is, then it should be easy to determine what is the outer function. The derivative of any composition of functions (when a function has another function inside of it) is given by h'(x) = g'(x)f'(g(x)). In the given example, the derivative would be h'(x) = (4x + 3)[3(2x2 + 3x)2], or (12x + 9)(2x2 + 3x)2 since g(x) = 2x2 + 3x and f(x) = x3.

2. What is dxn/dx?

Again, this is just testing the knowledge of a rudimentary rule given above.

3. Given f(x) = √(1-x2) find and simplify f'(x).

Well, this is basically just applying the rules of differentiating I reviewed above in a very straight forward manner. You must be able to recognize that you must use the chain rule with the inner function g(x) = 1 - x2, while the outer function f(x) = √x. Then the derivative becomes f'(x) = -2x / 2 √(1-x2), which simplifies to f'(x) = -x / √(1-x2).4. Find the equation of the line tangent to the graph of 8xy2 = (x + y)4 at the point (1/2, 1/2).

This was a rather tricky question, but I remember first determining how I forgot to take the derivative of x + y as 1 + dy/dx, instead of justdy/dx, and I also forgot to distribute the 8 through the left side and didn't read the question and gave him only a slope for an answer. I really need to take my time answering these questions =/. Well anyways...

The solution is shown above, but the main idea behind solving this question is just realizing that you must differentiate the equation implicitly, since y might represent any one of a variety of possible functions. If you can see that you must differentiate implicitly, it's pretty just grunt work applying the above rules and keeping in mind the fact that any time you differentiate the variable y, you must use the chain rule since it has some unknown buried inside of it, meaning that that term must be multiplied by y'. In the work you can see this being done. Say a function x + y = 4 was to be differentiated, then afterwards it would transform into 1 + 1y' = 0, or y' = -1.

Don't forget, as I did, that the question is asking for the equation to the tangent line at (1/2, 1/2), not just the slope of that tangent line. So, once you can solve for y' as shown above through implicit differentiation, you can use the point given (1/2, 1/2) to find the equation of the tangent line, thus completing the question.

5. (a) If f(x) = √(x2 + 9), use the limit definition of the derivative to find f'(0). You must show all work and use the limit definition properly to receive any credit.

The work for this question was once again fully shown by Mr. K, but the key to this question is realizing that it's asking for the derivative at a point, and not the derivative function of f(x). Using this knowledge, we can apply the limit definition at f(0) instead of f(x), and f(0+h) instead of at f(x+h). What Mr. K did with his work was that he determined what f(0) was by simply plugging 0 into x inside f(x), and then determined what f(0+h) would be by plugging in (0 + h) into f(x). He did this so that he could use the values he arrived at in these initial steps, and simply apply them to the limit definition as shown near the bottom of the slide.

5 (b) If g(x) = x2 - 3x + 4, use the definition of the derivative to find an equation or formula for the derivative of x. Again no credit will be awarded unless you demonstrate competent use of the limit definition and show all your work.

Once again you can see the solution to this question in the slide above. What you had to realize to solve this question is that you have to know what g(x+h) will be when expanded, as Mr. K has shown at the top portion of the slide. Once this is known, you must plug it into the fundamental limit definition of the derivative and things should start to cancel quickly and elegantly. In the final steps of the solution, once you plug in 0 into h, then the lim part goes away and any term with an h in it leaves along with it. Thus leaving 2x - 3, which gives the derivative of g(x), or rather gives g'(x).

Well that's it for my scribe post, I think that this might be one of my shortest scribe posts since way back in grade 12 pre-cal ^^. Too bad I wasn't present for the entire class, but I guess I tried to cover everything to the best of my ability since I was only present for maybe 20 minutes. Don't forgot to study tonight folks, since the test is tomorrow. I wish everyone the best of luck!Oh yes, almost forgot, the next scribe will be: John D.

MrSiwWy here for my second bob in intro to calculus. Oh wow, I just got home and I still have to scribe tonight =/ I don't expect to be faced with a terribly difficult test tomorrow, since we've already covered this unit in great depth in our AP calc class. Though, I must make sure to take my time this time and not to rush during any portion of the test, as I did on the pre-test (which cost me 2 rather easy marks). I'll probably explain that in more detail in my scribe post later tonight, but with that said, I don't think that there's anything I'm too worried about now. I usually make a longer bob than this, but I'll probably just integrate what I would usually say in my scribe post. Good luck on the test everyone ! Don't forget to practice those derivations!

Well, this unit about derivatives isn't too difficult considering that i learned this before. The only thing that i need to be careful on is working too fast for my own good. I usually make most of my mistakes on being too overconfident, but just to make sure I'll go over the orange and blue book tonight. Hopefully, i manage to get a good mark. Good luck everyone and study hard.

Tomorrow is the test! Wow it's quite surprising how so many people have dropped this course throughout the minor portion of what has passed. Well, tomorrow is the test on Derivatives! It shouldn't be too difficult to be honest, especially for the AP Calc Students who have already learned this unit. Though I understand the most part of this unit, I know that I may still stumble somewhere, as questions are always different: there are always new wrinkles that you can run into.

Summary of Differentiation:

Power Rule

Product Rule

F'(x) = f'(x) g(x) + f(x) g'(x)

Quotient Rule

Low D High, minus High D Low, All over Low Low (The SONG!)

F'(x) = [ g(x) f'(x) - f(x) g'(x) ] / [ g(x) ]^2

Chain Rule

F'(x) = f'( g(x) ) g'(x)

Including these rules, remember to understand the Definition of a Derivative and how to find the derivative using this definition.

Monday, April 14, 2008

Well I dropped the ball here, haven't really been paying too much attention to this particular blog of late. So by Murphy's law I was picked as scribe. Alright this isn't my best work I can tell you now, but this is how those two classes I have to scribe for go:

Day 1:

The point of this class was to come up with the equations of tangent lines and normal lines.-Tangent lines are lines touch that touch the graph only once on a certain interval (the graph may wiggle and the tangent line can cross the graph at another point that does not matter)-Normal lines are lines that are perpendicular to a tangent line.-Recall that the first derivative of any funtion gives the slope of the graph at any point-To find the equation of a tangent line at a point on a graph you need a couple things-You need a point and you need slope-You may use slope-intercept formula if you are given the y intercept y=mx+b-But you'll probably be using point-slope formula (y-y1) = m(x - x1)-So the general method to finding the formula of a tangent line to a point goes something like this: -Find the derivative of your function -Evaluate at your point -Plug it into point-slope formulaand presto! There's your answer!

To find the equation of a normal line (a line that is perpendicular to a tangent line) there is only one more step. When you find the slope you just need to take the negative reciprocal of that and then plug it into our formula.

The last question we had that class asks us to find when the tangent line is horizontal. Well that is when the derivative equals zero. So you come up with the formula of the derivative then solve for the zeroes. Then you plug those x-co's back into the original function to find the points where the tangent line equals zero.

Day 2:

This class is a little more complicated.-Mr.K started off with a talk about questions that the answers to were that just because we don't know for sure doesn't mean that something is or isn't there or happening.-The shadow of the balcony, we would infer that there is a balcony casting that shadow, doesn't have to be, could be something that looks like a balcony.-Then is that roof insulated? Well there is melted snow in places so we could assume no.-Then do those people agree with one another? Through the story of Mr.K since their body language is all somewhat the same they probably are, but we don't know for sure.

-We then found the derivative of a semicircle.-Mr.K showed us that we could define this function in other ways to get an infinite amount of other circle bits-and-pieces functions (see the slides)-This was to show us that there could be many functions buried within another and that even though we don't know which one there may be we have to treat it as if there is another function within it and therefor when we differentiate we have to use the chain rule.-One thing to notice about this is that the derivative might be in terms of y and x so you may need a set of coordinates to solve for the slope at a certain point instead of just an x-co.

Bob

I feel fairly confident for the upcoming test. No worries at all really. Schedule should die down a little so I wont' miss any more scribes or bobs!

G'night! (might edit this sometime to add some examples but I am *really* tired right now so sleep is a must... after bio...)

Monday, April 7, 2008

Hey everyone! I apologize for the super super late scribe post. I know you are all waiting for it so here it is.

March 19.The class started off with a discussion about Article 13 and the students' comments on their class blog. Moving on from that, we reviewed the sum/difference rule and then moved on to the new topic. The class topic was mostly about the product rule. All the information about the class work and the rule and how to apply it is in the slides below. Also we learned about the quotient rule but we ran out of time so it was continued in the other class.

March 28.This was our last class before the spring break. During this class we reviewed the quotient rule and Mr K. taught the class the quotient rule song. The song goes like this: "high de low minus low de high all over low low" which translate to: [f(x)^1 * g(x) - g(x)^1 * f(x)]/g(x)^2.But this class' real topic was... an introduction to the chain rule! The chain rule basically states that the whole function is called F(x). This F(x) is composed but different functions which are f(x) and g(x) which g(x) is the inner function and f(x) is the outer function. If examples are need it is in the slides below.

Thank you for your time and I do apologize for the super super late blog post. Now that I finished this scribe post the next scribe shall be... GREY-M.